Math Chapter 2 Review Key

Advanced Math Midterm Review Name: Key! Chapter 2 Find the distance between the set of points. 1. (6, 4) and (-8, 11) d = √((-8-6)² + (11-4)²) d = √196 + 49 d = √245 = √49 * √5 = 7√5 2. (40, 32) and (36, 20) d = √((40-36)² + (32-20)²) d = √16 + 144 d = √160 = √16 * √10 = 4√10 Find the midpoint of the set of points. 3. (6, 4) and (-8, 11) M = ((6-8)/2, (4+11)/2) M = (-1, 7.5) 4. (40, 32) and (36, 20) M = ((40+36)/2, (32+20)/2) M = (38, 26) 5. (12, -2) and (7, 2) M = ((12+7)/2, (-2+2)/2) M = (9.5, 0) 6. (5, 3) and (-6, 9) M = ((5-6)/2, (3+9)/2) M = (-1/2, 6) State whether the equation defines y as a function of x. 7. 2x + 4y = 8 Yes, function 8. y = x² Yes, function 9. y = ±√x Not a function 10. y = |x + 3| Yes, function State whether the set of ordered pairs defines y as a function of x. 11. (5, 2), (3, 7), (4, -3), (5, 6) Not a function 12. (6, 3), (-7, 3), (8, 3), (9, 3) Yes, function (OK for y values to repeat!) Evaluate the function. 13. Given f(x) = √(x² + 5) a. f(-2) = √((-2)² + 5) = √4 + 5 = √9 = 3 b. f(0) = √(0² + 5) = √5 c. f(3) = √(3² + 5) = √9 + 5 = √14 14. Given f(x) = 3x² + 1 a. f(-2) = 3(-2)² + 1 = 3 * 4 + 1 = 13 b. f(0) = 3(0)² + 1 = 1 c. f(3) = 3(3)² + 1 = 27 + 1 = 28 Write each interval as an inequality and graph on a number line. 15. -2 < x ≤ 5 (-2, 5] 16. x > 3 (3, ∞) 17. x ≤ 6 (-∞, 6] 18. -1 ≤ x ≤ 1 [-1, 1] Write each inequality in interval notation. 19. -4 < x ≤ 1 (-4, 1] 20. x < 3 or x ≥ 6 (-∞, 3) U [6, ∞) 21. -8 < x < -2 (-8, -2) 22. -3 ≤ x ≤ 2 [-3, 2] 23. x < 6 (-∞, 6) 24. x > -5 (-5, ∞) Determine the domain of the function represented by the given equation. Write your answer in INTERVAL NOTATION. 25. f(x) = 2x + 5 (-∞, ∞) 26. f(x) = x² - 5x (-∞, ∞) 27. f(x) = √(6 + x) 6 + x ≥ 0 x ≥ -6 [-6, ∞) 28. f(x) = √(x - 3) x - 3 ≥ 0 x ≥ 3 [3, ∞) 29. f(x) = 4 / (x - 6) x - 6 ≠ 0 x ≠ 6 (-∞, 6) U (6, ∞) 30. f(x) = 7 / (x + 8) x + 8 ≠ 0 x ≠ -8 (-∞, -8) U (-8, ∞) Find the real value or values of a in the domain of f for which f(a) equals the given number. 31. f(x) = 3x - 2; f(a) = 10 10 = 3a - 2 12 = 3a a = 4 32. f(x) = 2 - 5x; f(a) = 7 7 = 2 - 5a 5 = -5a a = -1 33. f(x) = x² + 2x - 2; f(a) = 1 1 = a² + 2a - 2 0 = a² + 2a - 3 0 = (a + 3)(a - 1) a = -3, 1 34. f(x) = |x + 2|; f(a) = 6 6 = |a + 2| 6 = a + 2 or -6 = a + 2 a = 4 or a = -8 Find the zeros, or x-intercepts, of f(x). 35. f(x) = 3x + 2 0 = 3x + 2 -2 = 3x -2/3 = x 36. f(x) = 6 + 2x 0 = 6 + 2x -6 = 2x -3 = x 37. f(x) = x² - 5x - 24 0 = (x - 8)(x + 3) x - 8 = 0 or x + 3 = 0 x = 8, -3 38. f(x) = x² + 4x - 21 0 = (x + 7)(x - 3) x + 7 = 0 or x - 3 = 0 x = -7, 3 Find the slope through the given points. 39. (8, -2) and (-4, 10) m = (10 - (-2))/(-4 - 8) = 12/-12 = -1 40. (-2, -7) and (4, -9) m = (-9 - (-7))/(4 - (-2)) = -2/6 = -1/3 Graph y as a function of x by finding the slope and y-intercept. 41. y = -3x + 5 Slope: -3 y-intercept: 5 42. y = 4x - 2 Slope: 4 y-intercept: -2 43. 2x - 3y = 12 -3y = -2x + 12 y = 2/3x - 4 Slope: 2/3 y-intercept: -4 44. -3x + 5y = -15 5y = 3x - 15 y = 3/5x - 3 Slope: 3/5 y-intercept: -3 Find the equation of the indicated line in slope-intercept form. 45. Slope: -17, y-intercept: (0, -5) y = -17x - 5 46. Slope: 6, y-intercept: (0, 2) y = 6x + 2 47. Slope: 12, Through (4, -6) -6 = 12(4) + b -6 = 48 + b -54 = b y = 12x - 54 48. Slope: -3, Through (-5, -1) -1 = -3(-5) + b -1 = 15 + b -16 = b y = -3x - 16 49. Through (5, 6) and (2, -8) m = (-8 - 6)/(2 - 5) = -14/-3 = 14/3 -8 = (14/3)(2) + b -8 = 28/3 + b -52/3 = b y = (14/3)x - 52/3 50. Through (-5, 6) and (-3, -4) m = (-4 - 6)/(-3 - (-5)) = -10/2 = -5 -4 = -5(-3) + b -4 = 15 + b -19 = b y = -5x - 19 Find the equation of the line, in slope-intercept form, that satisfies the given conditions. 51. The graph is parallel to the graph of y = (2/3)x + 3 and passes through the point whose coordinates are (-2, 1). m = 2/3 1 = (2/3)(-2) + b 1 = -4/3 + b 7/3 = b y = (2/3)x + 7/3 52. The graph is parallel to the graph of y = 3x - 1 and passes through the point whose coordinates are (-5, 4). m = 3 4 = 3(-5) + b 4 = -15 + b 19 = b y = 3x + 19 53. The graph is perpendicular to the graph of 3x - 2y = 5 and passes through the point whose coordinates are (-3, 4). 3x - 2y = 5 -2y = -3x + 5 y = (3/2)x - 5/2 m = -2/3 4 = (-2/3)(-3) + b 4 = 2 + b 2 = b y = (-2/3)x + 2 54. The graph is perpendicular to the graph of y = -x + 3 and passes through the point whose coordinates are (-5, 2). m = -1 m = 1 2 = 1(-5) + b 2 = -5 + b 7 = b y = x + 7